Mister Exam
x2-y2=32; x2+y2=40
The solution
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x2 - y2 = 32
$$x_{2} - y_{2} = 32$$
x2 + y2 = 40
$$x_{2} + y_{2} = 40$$
x2 + y2 = 40
Detail solution
Given the system of equations
$$x_{2} - y_{2} = 32$$
$$x_{2} + y_{2} = 40$$
Let's express from equation 1 x2
$$x_{2} - y_{2} = 32$$
Let's move the summand with the variable y2 from the left part to the right part performing the sign change
$$x_{2} = y_{2} + 32$$
$$x_{2} = y_{2} + 32$$
Let's try the obtained element x2 to 2-th equation
$$x_{2} + y_{2} = 40$$
We get:
$$y_{2} + \left(y_{2} + 32\right) = 40$$
$$2 y_{2} + 32 = 40$$
We move the free summand 32 from the left part to the right part performing the sign change
$$2 y_{2} = -32 + 40$$
$$2 y_{2} = 8$$
Let's divide both parts of the equation by the multiplier of y2
$$\frac{2 y_{2}}{2} = \frac{8}{2}$$
$$y_{2} = 4$$
Because
$$x_{2} = y_{2} + 32$$
then
$$x_{2} = 4 + 32$$
$$x_{2} = 36$$
The answer:
$$x_{2} = 36$$
$$y_{2} = 4$$
$$x_{2} - y_{2} = 32$$
$$x_{2} + y_{2} = 40$$
Let's express from equation 1 x2
$$x_{2} - y_{2} = 32$$
Let's move the summand with the variable y2 from the left part to the right part performing the sign change
$$x_{2} = y_{2} + 32$$
$$x_{2} = y_{2} + 32$$
Let's try the obtained element x2 to 2-th equation
$$x_{2} + y_{2} = 40$$
We get:
$$y_{2} + \left(y_{2} + 32\right) = 40$$
$$2 y_{2} + 32 = 40$$
We move the free summand 32 from the left part to the right part performing the sign change
$$2 y_{2} = -32 + 40$$
$$2 y_{2} = 8$$
Let's divide both parts of the equation by the multiplier of y2
$$\frac{2 y_{2}}{2} = \frac{8}{2}$$
$$y_{2} = 4$$
Because
$$x_{2} = y_{2} + 32$$
then
$$x_{2} = 4 + 32$$
$$x_{2} = 36$$
The answer:
$$x_{2} = 36$$
$$y_{2} = 4$$
Rapid solution
$$x_{21} = 36$$
=
$$36$$
=
$$y_{21} = 4$$
=
$$4$$
=
=
$$36$$
=
36
$$y_{21} = 4$$
=
$$4$$
=
4
Gaussian elimination
Given the system of equations
$$x_{2} - y_{2} = 32$$
$$x_{2} + y_{2} = 40$$
We give the system of equations to the canonical form
$$x_{2} - y_{2} = 32$$
$$x_{2} + y_{2} = 40$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & -1 & 32\\1 & 1 & 40\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & -1 & 32\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-1 + 1 & 1 - -1 & \left(-1\right) 32 + 40\end{matrix}\right] = \left[\begin{matrix}0 & 2 & 8\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & -1 & 32\\0 & 2 & 8\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-1\\2\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 2 & 8\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-1\right) 0}{2} & -1 - \frac{\left(-1\right) 2}{2} & 32 - \frac{\left(-1\right) 8}{2}\end{matrix}\right] = \left[\begin{matrix}1 & 0 & 36\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 36\\0 & 2 & 8\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} - 36 = 0$$
$$2 x_{2} - 8 = 0$$
We get the answer:
$$x_{1} = 36$$
$$x_{2} = 4$$
$$x_{2} - y_{2} = 32$$
$$x_{2} + y_{2} = 40$$
We give the system of equations to the canonical form
$$x_{2} - y_{2} = 32$$
$$x_{2} + y_{2} = 40$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & -1 & 32\\1 & 1 & 40\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & -1 & 32\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-1 + 1 & 1 - -1 & \left(-1\right) 32 + 40\end{matrix}\right] = \left[\begin{matrix}0 & 2 & 8\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & -1 & 32\\0 & 2 & 8\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-1\\2\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 2 & 8\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-1\right) 0}{2} & -1 - \frac{\left(-1\right) 2}{2} & 32 - \frac{\left(-1\right) 8}{2}\end{matrix}\right] = \left[\begin{matrix}1 & 0 & 36\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 36\\0 & 2 & 8\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} - 36 = 0$$
$$2 x_{2} - 8 = 0$$
We get the answer:
$$x_{1} = 36$$
$$x_{2} = 4$$
Cramer's rule
$$x_{2} - y_{2} = 32$$
$$x_{2} + y_{2} = 40$$
We give the system of equations to the canonical form
$$x_{2} - y_{2} = 32$$
$$x_{2} + y_{2} = 40$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} - x_{2}\\x_{1} + x_{2}\end{matrix}\right] = \left[\begin{matrix}32\\40\end{matrix}\right]$$
- this is the system of equations that has the form
A*x = B
Let´s find a solution of this matrix equations using Cramer´s rule:
Since the determinant of the matrix:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & -1\\1 & 1\end{matrix}\right] \right)} = 2$$
, then
The root xi is obtained by dividing the determinant of the matrix Ai. by the determinant of the matrix A.
( Ai we get it by replacement in the matrix A i-th column with column B )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}32 & -1\\40 & 1\end{matrix}\right] \right)}}{2} = 36$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 32\\1 & 40\end{matrix}\right] \right)}}{2} = 4$$
$$x_{2} + y_{2} = 40$$
We give the system of equations to the canonical form
$$x_{2} - y_{2} = 32$$
$$x_{2} + y_{2} = 40$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} - x_{2}\\x_{1} + x_{2}\end{matrix}\right] = \left[\begin{matrix}32\\40\end{matrix}\right]$$
- this is the system of equations that has the form
A*x = B
Let´s find a solution of this matrix equations using Cramer´s rule:
Since the determinant of the matrix:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & -1\\1 & 1\end{matrix}\right] \right)} = 2$$
, then
The root xi is obtained by dividing the determinant of the matrix Ai. by the determinant of the matrix A.
( Ai we get it by replacement in the matrix A i-th column with column B )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}32 & -1\\40 & 1\end{matrix}\right] \right)}}{2} = 36$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 32\\1 & 40\end{matrix}\right] \right)}}{2} = 4$$